# Work Required

1. Feb 8, 2009

### shimizua

1. The problem statement, all variables and given/known data
Calculate the work required to move a charge of -0.51x10^-12 C from i' to b'.
http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype54/prob04a_threeqcontour.gif

2. Relevant equations
i know that the equation is -W=qEl and that it can also be expressed as -W=-q$$\Delta$$V or W=q$$\Delta$$V

3. The attempt at a solution
well q is given as -0.51x10^-12
the problem i am having is how do i find $$\Delta$$V without knowing E? cause i know i can figure out l just by using the picture. So if you can just help me with find out how to get $$\Delta$$V that would be great!

2. Feb 8, 2009

### LowlyPion

Looks like your ΔV is going from +3 to -3. That looks like a Voltage drop of 6v from the image, or ΔV = -6.

Just count the contours is my method.

3. Feb 8, 2009

### Redbelly98

Staff Emeritus
The figure is a contour plot of the potential. Each contour line represents a constant potential value.

The figure also indicates which contour lines correspond to -5V, 0V, and +5V. Using that, can you say:
What is the voltage at point i?
What is the voltage at point b?

4. Feb 9, 2009

### shimizua

ok so i did the (-.51x10^-12)(-6) and got an answer of 3.06x10^-12 J and it was said to be wrong. Where did i got wrong?

5. Feb 9, 2009

### shimizua

oh and i was given this hint when i put my answer in

The equipotential lines shown are separated by 1 kV. (NOTE! That's kV, not volts!) Work to move a charge is the increase in potential energy of the charge. The potential energy is the potential times the charge.

6. Feb 9, 2009

### cjl

The answer remains the same though, whether it is kV or V. It's still a matter of counting contour lines.

7. Feb 9, 2009

### Redbelly98

Staff Emeritus
If it's actually 1 kV, and not 1 V, for the contour intervals, then instead of 6V it is really ____?

8. Feb 9, 2009

### shimizua

ok so then it would mean that it was really 6 kV or 6000 V and then when i did that i got 3x10^-9 J and that also came up wrong

9. Feb 9, 2009

### Newton V

your graph is ill-made....

10. Feb 9, 2009

### shimizua

haha, it is the graph that they gave up. i didnt make it

11. Feb 9, 2009

### Newton V

ok i give up... i am having my own probelms...

12. Feb 9, 2009

### LowlyPion

It's plus Work moving an opposite charge away from + and toward a - V.

If the contours are 103v then it's 3.06 * 10-9 J or 306 ergs.

If it's not precision or units then I've no idea what is asked.

13. Feb 10, 2009

### shimizua

i got the answer. i did the 3.06e-09 J and it seemed to work even though i did 3.1e-09 before. i guess it had to be 06. thanks for all your help guys

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